# Math Help - Simple quartile problem.

1. ## Simple quartile problem.

A density curve consists of a straight line segment that begins at the origin (0,0) and has a slop 1.

b) determine the median, first quartile, and third

c) relative to the median, where would you expect the mean of the distribution?

d) what percent of the observations lie below .5? above 1.5?

I have the answers but i do not know how to get them. Can someone show me how there problems work?

2. It's a triangle, right?

$
\frac{1}{2} \cdot Base \cdot Height\;=\;Area
$

$
\frac{1}{2} \cdot x \cdot x\;=\;1
$

Solve for 'x' and find out where the distribution ends.

All else should be relatively clear from there.

Note: Are you serious about "c"? That can't be typed correctly.

3. Originally Posted by TKHunny
It's a triangle, right?

$
\frac{1}{2} \cdot Base \cdot Height\;=\;Area
$

$
\frac{1}{2} \cdot x \cdot x\;=\;1
$

Solve for 'x' and find out where the distribution ends.

All else should be relatively clear from there.

Note: Are you serious about "c"? That can't be typed correctly.

oh yeah my bad it's it's median. How can I get the answer.

And so for Q1 it is (X^2)/2 = .25 . Therefore the answer should be .707? Am I rignt?

A density curve consists of a straight line segment that begins at the origin (0,0) and has a slop 1.

b) determine the median, first quartile, and third

c) relative to the median, where would you expect the mean of the distribution?

d) what percent of the observations lie below .5? above 1.5?

I have the answers but i do not know how to get them. Can someone show me how there problems work?
By density curve do you mean probability density function (pdf). If so, then the rule must be:

$f(x) = \left \{ \begin{array}{cc}
x & 0 \leq x \leq \sqrt{2} \\
0 & \text{otherwise}\end{array} \right.$